68.1.7. Small Signal Parameters Calculation

So, let me summarize the operating point.

So, we do have I = I = 2 mA, V = V = 6 V. And yeah so, the other things we already have obtained namely I = 20 µA and I = 10 µA right. So, since this is 6 V and this is point 6 V, this is supply voltage is 12 V, here the voltage it is 12 ‒ points 0.6. So, that is 11.4 V. So, that is the operating point.

This chunk summarizes the operating points of two transistors in the circuit. The collector currents (I) through both transistors are equal at 2 mA. The voltages across the transistors (V) are also equal at 6 V. The base currents (I1 and I2) are 20 µA and 10 µA, respectively. The reference to the supply voltage being 12 V minus the voltage drop across the transistor (0.6 V) shows how the values relate to the overall circuit design.

Think of the operating point like a car's fuel gauge. Just as you want to know how much fuel you have (I1, I2) and how full your tank is (supply voltage), in circuits, you need to know the currents through various components and their voltage levels to ensure they function properly.

So, from that we can calculate the small signal parameters of the transistors namely in a g say g of transistor-1 it is thermal equivalent voltage we can consider that is 26 mV. So, this is ℧. So, likewise g it is also = ℧, then r = β of transistor-1 divided by g of the transistor.

So, that is equal to 100 × 13 so that = 1.3 kΩ. So, likewise you can also find that r which is β 200 and then g it is . So, that = 2.6 KΩ.

In this part, the small signal parameters for the transistors are calculated. The transconductance (gm) is given as the thermal equivalent voltage of 26 mV. The output resistance (r) for transistor-1 is calculated using the formula r = β/gm. This means for transistor-1 with a β of 100, r becomes 1.3 kΩ. For transistor-2, using a β of 200, it yields a resistance of 2.6 kΩ.

Imagine trying to push a swing (transistor). The ease you find in pushing it is like the small signal parameters. The force you apply corresponds to the input (gm), and how far the swing moves relates to the output resistance (r). A swing with a lighter frame will move more easily, akin to a smaller output resistance allowing greater movement in response to input.

Now next thing is that, how do we find the small signal gain and so and so, so in the next slide what we can do we are going to draw the small signal equivalent circuit, but to calculate the gain we need to remember these parameters particularly small signal parameters values. So, you have to keep that in mind and then we will be utilizing this parameter value in the calculation of small signal gain and so and so.

This section transitions to calculating the small signal gain of the circuit. It emphasizes the importance of small signal parameters established earlier. These parameters will be crucial for determining how well the transistors amplify small signals from the input to the output.

Consider a public address system. Suppose you want to ensure the microphone captures every soft whisper effectively. The microphone's sensitivity and the amplifier's configurations (analogous to small signal parameters) will determine how well those whispers are projected through the speakers. The better your settings, the clearer your sound quality.

So, to find the output resistance what we have to do? We can stimulate this circuit by a signal called v and then we can observe the corresponding current here. If the current is i then you can say that the output resistance R = . So, that is the methodology we follow to find the port resistance and while we are doing this exercise we have to consider the other sources signal sources at 0.

To find the output resistance of the circuit, the method involves applying a test voltage (v) and measuring the resulting current (i). This allows calculation of the output resistance using Ohm's law: R = V/I. It’s important to ensure that other signals or sources are set to zero during this process to avoid interference that could skew results.

Think of an elastic band. When you pull on one end (applying a signal), you can observe how much force it takes to stretch it (measuring current). This resistance to stretching is like the output resistance you're calculating, and you wouldn’t want other people pulling on the band at the same time, as that would give you mixed results.

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